Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:

import java.io.*;
import java.util.*;

public class C {

    static StreamTokenizer st;
    static PrintWriter out;

    public static void main(String[] args) throws IOException {
        st = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        out = new PrintWriter(System.out);
        int n = ri();
        int m = ri();
        boolean rot = false;
        if (n > m) {
            rot = true;
            int t = n;
            n = m;
            m = t;
        }
        int[][] a = new int[n][m];
        int[] min = new int[n];
        if (!rot) {
            for (int i = 0; i < n; i++) {
                int mn = Integer.MAX_VALUE;
                for (int j = 0; j < m; j++) {
                    mn = Integer.min(a[i][j] = ri(), mn);
                }
                min[i] = mn;
            }
        } else {
            for (int i = 0; i < m; i++) {
                int mn = Integer.MAX_VALUE;
                for (int j = 0; j < n; j++) {
                    a[j][i] = ri();
                }
            }
            for (int i = 0; i < n; i++) {
                min[i] = Integer.MAX_VALUE;
                for (int j = 0; j < m; j++) {
                    min[i] = Integer.min(a[i][j], min[i]);
                }
            }
        }
        StringBuilder res = new StringBuilder();
        int steps = 0;
        String s1 = rot ? "col " : "row ";
        String s2 = rot ? "row " : "col ";
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < min[i]; j++) {
                res.append(s1);
                res.append((i + 1));
                res.append(System.lineSeparator());
            }
            steps += min[i];
            for (int j = 0; j < m; j++) {
                a[i][j] -= min[i];
            }
        }

        boolean done = true;
        for (int col = 0; col < m && done; col++) {
            for (int row = 1; row < n; row++) {
                if (a[row][col] != a[0][col]) {
                    done = false;
                    break;
                }
            }
            if (done) {
                for (int i = 0; i < a[0][col]; i++) {
                    res.append(s2);
                    res.append((col + 1));
                    res.append(System.lineSeparator());
                }
                steps += a[0][col];
            }
        }
        if (done) {
            out.println(steps);
            out.println(res.toString());
        } else {
            out.println(-1);
        }
        out.close();
    }

    static int ri() throws IOException {
        st.nextToken();
        return (int) st.nval;
    }

}